Coin Change II

Statement#

Suppose you are given a list of coins and a certain amount of money. Each coin in the list is unique and of a different denomination. You are required to count the number of ways the provided coins can sum up to represent the given amount. If there is no possible way to represent that amount, then return 0.

Note: You may assume that for each combination you make, you have an infinite number of each coin. In simpler terms, you can use a specific coin as many times as you want.

Let's say you have only two coins, $10$ and $20$ cents, and you want to represent the total amount of $30$ cents using these coins. There are only two ways to do this, you can use either of the following combinations:

• 3 coins of $10$ cents: $10+10+10=30$.

• 1 coin of $10$ cents and 1 coin of $20$ cents: $10+20=30.$

Constraints:

• 1 <= coins.length <= 300

• 1 <= coins[i] <= 5000

• All the coins have a unique value.

• 0 <= amount <= 5000

Examples#

Let's see a few more examples to get a better understanding of the problem statement:

Try it yourself#

Implement your solution in the following playground.

Note: If you clicked the “Submit” button and the code timed out, this means that your solution needs to be optimized in terms of running time.

Hint: Use dynamic programming and see the magic.

Solution#

We will first explore the naive recursive solution to this problem and then see how it can be improved using the Unbounded Knapsack dynamic programming pattern.

Naive approach#

A naive approach to solve this problem would be to make all the possible combinations of coins or generate all subsets of coin denominations that sum up to the required amount.

While making the combinations, a point to keep in mind is that we should try to avoid repeatedly counting the same combinations. For example, $10+20$ cents add up to $30$ cents, and so do $20+10$. In the context of combinations, both these sequences are the same, and we will count them as one combination. We can achieve this by using a subset of coins every time we consider them for combinations. The first time, we can use all $n$ coins, the second time, we can use $n−1$ coins, that is, by excluding the largest one, and so on. This way, it is not possible to consider the same denomination more than once.

As a base case, we return $1$ when the target amount is zero because there is only one way to represent zero irrespective of the number and kinds of coins available. Similarly, for the second base case, if at any point during our search for combinations, the remaining value needed to reach the total amount becomes less than zero, we return $0$.

The slides below will help you visualize the process.

Now, let's look at the code in the code widget below for the solution we just discussed.

Note: Please observe that if you include the test case commented out in the driver code, the solution is likely to time out. Try to solve the larger problem using the dynamic programming solutions provided below and see the difference.

Time complexity#

If the length of the list coins is $n$, and the amount to be made out of them is $C$, the time complexity of the recursive algorithm would be $O(n^C)$. This is because the maximum height of the call tree is $C$, and there may be at most $n$ branches from each node in the call tree. Therefore, the total number of nodes in such a tree is bound by $n^C$. Therefore, the time complexity would be $O(n^C)$.

Space complexity#

The space complexity for this algorithm is $O(C)$.

Optimized solution using dynamic programming#

As the recursive solution to this problem is very costly, let's see if we can reduce this cost in any way. Dynamic programming helps us avoid recomputing the same subproblems. Now let's analyze our recursive solution to see if it has the properties needed for conversion to dynamic programming.

• Optimal substructure: If we wanted to find the solution to the problem of counting an amount, $C$, given $n$ different coins, and we knew the answer to the $n$ subproblems formed by subtracting each of the $n$ coins from the amount $C$, we could simply sum up the answer to these subproblems, and that would give us the answer to our original problem. Therefore, this problem obeys the optimal substructure property.

• Overlapping subproblems: The algorithm solves the same subproblems again and again, so it also has the second property of dynamic programming.

Top-down solution#

The top-down solution, commonly known as the memoization technique, is an enhancement of the recursive solution. It overcomes the problem of calculating solutions to subproblems over and over again by storing the subproblem solutions, and reusing the stored solutions when the same subproblem is encountered again. Therefore, the only addition to this algorithm compared to the simple recursive one is the addition of memoization. We store all the results in memo and then retrieve them as needed. Since we had two defining variables here, the amount, and the maximum value, we use them to uniquely identify each subproblem.

Note: Please observe that if you now uncomment the large test case in the driver code, the dynamic programming solution will still work.

Time complexity#

Let's see how many unique subproblems are possible that we can evaluate. The value of amount will not be greater than the one provided at the start of the algorithm, let’s call it $C$. Similarly, since we have $n$ number of different coins, the maximum variable will therefore be bounded by $n$. Therefore, we will have, at most $C\times n$ subproblems. Therefore, the time complexity to evaluate these problems would be $O(Cn)$.

Space complexity#

The space required to hold the results of these subproblems would be $C\times n$, so the space complexity is $O(Cn)$ as well.

Bottom-up solution#

Now let’s look at the bottom-up dynamic programming approach to solving this problem. The bottom-up solution, also known as the tabulation technique, is an iterative method of solving dynamic programming problems. To make up an amount using $n$ coins, we just need to iteratively count the ways in which we can either make the amount using the $n^{th}$ coin or without using it.

Here, we first create a table $dp$ of size $(C+1) \times (n)$, where $C$ is the target amount and $n$ is the number of coins we have. The row signifies the amount, $i$, at any given point, and the column signifies the values of the available coins, $j$. Each cell of our table, $dp[i][j]$, represents the number of ways to make the amount $i$ using coins $0,1, 2, ... j$.

First, as per the base case, for the amount $0$, $i=0$ , we add "$1$s" in the first row of the table. This is because there is always one way to make an amount of $0$ by using none of the available coins, that is, we choose not to pick any of the coins $1$, $2$, or $3$.

Next, we fill up each cell of our table using the unbounded knapsack dynamic programming pattern. In the pattern, for each cell, we need the sum of two values we get from two of our choices:

• Use the coin: We only use the coin if $i - coins[j] >= 0$. If we pick $coins[j]$, provided $i - coins[j] >= 0$, then the amount $i$ reduces by $coins[j]$. Now, we need the number of ways to make this remaining amount using coins $0, 1, 2, ...j$. We already have this value present at $dp[i - coins[j]][j]$.

• Don't use the coin: We don't use the coin if $i - coins[j] < 0$. If we don't pick $coins[j]$, given $i - coins[j] < 0$, then we have to figure out the number of ways to make the amount $i$ using only coins $0, 1, 2, ...j - 1$. This value is already present at $dp[i][j-1]$.

Let’s look at the following illustration to get a better understanding of the algorithm:

Let's have a look at the code for the bottom-up approach as well.

Note: Please observe that if you now uncomment the large test case in the driver code, the dynamic programming solution will still work.

Time complexity#

As apparent from the visualization as well, we are iterating over a table of size $C\times n$ , where $C$ is the target amount, and $n$ is the number of coins. Therefore, the time complexity of this algorithm is $O(Cn)$.

Space complexity#

The space complexity of this algorithm is $O(Cn)$.

Can we do better?#

You must have noticed in the above algorithm that to fill up a column, we only require the previous column’s values; that is, for filling the column against the $n^{th}$ coin, we require the values from the previous column representing the $(n-1)^{th}$ coin. Therefore, there is no point in storing all the previous $(n-2)$ columns.

We can further improve this approach by using a lookup array of length $amount + 1$ instead of creating a table. Next, we update this array for each coin using the same calculation which we used earlier.

The time complexity would remain the same, $O(Cn)$, because we still have to do the calculation for each value amount and each coin. The space complexity, however, reduces to $O( C )$ since we are only maintaining an array of the size $amount + 1$.